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How to evaluate the integral: $$\int_{0}^{\pi}\arctan\left(\frac{1+\cos x}{2+\sin x}\right)dx$$ I didn't get it by partial integration,Any help is appreciated.

I still can't solve this integral. That's a good question. Why doesn't anyone answer it?

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closed as off-topic by Namaste, mickep, user21820, Carl Mummert, user223391 Oct 22 '17 at 14:46

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  • $\begingroup$ Where did you see this integral? $\endgroup$ – pisco Jan 20 '18 at 9:07
  • $\begingroup$ @pisco125 I'm sorry. It's been a long time. I forgot. $\endgroup$ – JamesJ Jan 20 '18 at 9:28
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Through $\arctan(x)+\arctan(y)=\arctan\left(\frac{x+y}{1-xy}\right)$ and symmetry we get that the given integral equals $$ \int_{0}^{\pi/2}\arctan\left(\frac{2+\sin(x)}{2+2\sin(x)}\right)\,dx =\int_{0}^{1}\arctan\left(\frac{2+z}{2+2z}\right)\frac{dz}{\sqrt{1-z^2}}$$ hence it is enough to compute $$ \int_{3/4}^{1}\frac{\arctan(u)\,du}{(2u-1)\sqrt{4u-3}}\quad\text{or}\quad \int_{0}^{1}\arctan\left(\frac{3+v^2}{4}\right)\frac{dv}{1+v^2}.$$ This can be done through integration by parts and the dilogarithms machinery, but the final outcome is not really nice. At least, not the one I got, but maybe I missed some crucial simplification. Anyway, the approach just outlined tends to work pretty well for similar Ahmed-like integrals.

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  • $\begingroup$ I don't really understand this step? $\endgroup$ – JamesJ Oct 20 '17 at 5:47
  • $\begingroup$ @JamesJ: which step is troubling you? $\endgroup$ – Jack D'Aurizio Oct 20 '17 at 9:32
  • $\begingroup$ What's the connection between the first definite integral and my integral? $\endgroup$ – JamesJ Oct 20 '17 at 9:58
  • $\begingroup$ @JamesJ: they are the same. You integral equals $$\int_{0}^{\pi/2}\arctan\frac{1+\cos x}{2+\sin x}\,dx+\int_{0}^{\pi/2}\arctan\frac{1-\cos x}{2+\sin x}\,dx = \int_{0}^{\pi/2}\arctan\frac{2+\sin x}{2+2\sin x}\,dx$$ $\endgroup$ – Jack D'Aurizio Oct 20 '17 at 10:20
  • $\begingroup$ But that's what mine is:$\frac{1-\sin x}{2+\cos x}$? $\endgroup$ – JamesJ Oct 20 '17 at 10:35

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